How Do I Integrate x * sin(pi*x/a) * sin(2pi*x/a)?

In summary, to integrate x * sin(pi*x/a) * sin(2pi*x/a) over the interval 0 to a, one can use the trigonometric identity sin(2y) = 2sin(y)cos(y) to convert the integral to 2y * (sin(y))^2 * cos(y) dy. This can then be solved using integration by parts, setting u = 2y and dv = (sin(y))^2 * cos(y) dy. After solving for v, the integral can be solved using the standard approach, resulting in the answer 2xsin(x) - cos(x) - 3xcos(x)sin(x) + 3cos^2(x) + sin^
  • #1
mmwave
647
2
As a minor step in a quantum mechanics problem I need to integrate x * sin(pi*x/a) * sin (2pi*x/a) over the interval 0 to a.

I have had no luck and can't find it in my tables. Do I have to convert it to exponentials and integrate all those terms?

I tried using sin 2y = 2siny cosy to get
[inte] y * 2sin2(y) cos(y) dy
but that didn't get me anywhere. Help would be greatly appreciated.
 
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  • #2
Originally posted by mmwave

I tried using sin 2y = 2siny cosy to get [inte] 2sin^2 (y) cos (y) dy
but that didn't get me anywhere. Help would be greatly appreciated.
let u=siny
then du=cosydy
and [inte]2sin2y cosydy=2[inte]u2du=2/3u3
 
  • #3
That's not going to do it -- you overlooked the x term.

But,
xsin(x)sin(2x) = x*sin(x)*2sin(x)cos(x)
= 2xsin2(x)cos(x)
= 2x(1-cos2(x))*cos(x)
= 2xcos(x) -2xcos3(x)

and you can find
∫ 2xcos(x) dx
and
∫ 2xcos3(x) dx
using integration by parts.

Can you take it from there?
 
  • #4


Originally posted by StephenPrivitera
let u=siny
then du=cosydy
and [inte]2sin2y cosydy=2[inte]u2du=2/3u3

My fault, I left the y out of the integral you copied. But I did try this in combination with the y term and then integration by parts as
u = y, du = dy and dv = your u substitution above. It didn't seem to work.
 
  • #5
Originally posted by gnome
That's not going to do it -- you overlooked the x term.

But,
xsin(x)sin(2x) = x*sin(x)*2sin(x)cos(x)
= 2xsin2(x)cos(x)
= 2x(1-cos2(x))*cos(x)
= 2xcos(x) -2xcos3(x)

and you can find
∫ 2xcos(x) dx
and
∫ 2xcos3(x) dx
using integration by parts.

Can you take it from there?

I think I can I think I can ...

The advantage of the conversion to cosines is that I get a sum of terms instead of a product of sines?
 
  • #6
Let's take it from the top. Your application of a trigonometric identity was a good choice, because it makes all of the trig functions involved have the same argument.

∫ 2 y (sin y)2 cos y dy

The others have mentioned integration by parts, and, well, this is a prototypical application of IBP since we have y multiplied by something we can integrate!

So we do IBP, setting:

u = 2y
dv = (sin y)2 cos y dy

du = 2 dy
v = ∫ (sin y)2 cos y dy


Stephen has given the standard approach on how to perform this subintegral. Applying the substitution will allow you to integrate and compute what v is supposed to be.

Can you take it from here?
 
  • #7
is the integral of 2xcosx dx :

2xsinx + cosx + c?


and integral of 2x(cosx)^3 dx:

xcos^4/2 - cos^5/20 + c?
 
  • #8
I got it thanks. Then I got lots of practice in variations of the same theme. Then I got to do lots of integrals with x2. These were much tougher.

Does anyone know a book with higher powers of sin and cosine terms?

I like Alan Jefferey's book but the tables stop at

[inte] x sin3x dx and
[inte] x2 sin2x dx

In my problems I'm getting sin and cosines to the 4,5 and 6th powers.
 
  • #9
Here's what I got, 2 answers, don't know if anyone of them is right.


2[xsinx - cosx - 3xcosxsinx + 3cos^2(x) - 3sin^2(x)] + C

where x = pi*x/a


2(xsinx - cosx - 3xcosxsinx + 3cos^2(x) + sin^2(x)) + C

where x = pi*x/a
 

1. What is the general formula for the integral of X * sin x * sin2x?

The general formula for the integral of X * sin x * sin2x is ∫X * sin x * sin2x dx = -X * (1/4) * cos3x + C

2. How do you solve the integral of X * sin x * sin2x?

To solve the integral of X * sin x * sin2x, you can use integration by parts or trigonometric substitution.

3. What are the limits of integration for the integral of X * sin x * sin2x?

The limits of integration for the integral of X * sin x * sin2x can vary depending on the given problem. Generally, they will be given in the original problem or can be determined by the context of the problem.

4. Can the integral of X * sin x * sin2x be evaluated without using integration techniques?

No, the integral of X * sin x * sin2x cannot be evaluated without using integration techniques. It involves a variable coefficient and the product of two trigonometric functions, making it necessary to use integration techniques to solve.

5. How is the integral of X * sin x * sin2x used in real-life applications?

The integral of X * sin x * sin2x has various applications in physics, engineering, and other sciences. It can be used, for example, in calculating the work done by a force acting on a moving object or in determining the energy of a simple harmonic oscillator.

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